Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <cassert>
#include <set>
#include <queue>
using namespace std;
using ll = long long;
const int INF = 1<<30;
const int MOD = 1e9 + 7;
const int MAX_N = 1e5 + 10;
int N, M, L;
set<int> T[MAX_N];
vector<int> scnt, tcnt, sdist, tdist;
void search(int s, int t, bool flag)
{
    queue<int> que;
    vector<int> dist(N, INF);
    vector<int> cnt(N, 0);
    dist[s] = 0;
    que.push(s);
    while(que.size())
    {
        int v = que.front(); que.pop();
        for(auto u : T[v])
        {
            if(dist[u] > dist[v] + 1)
            {
                dist[u] = dist[v] + 1;
                que.push(u);
            }
        }
    }
    for(int i = 0; i < N; i++)
    {
        if(dist[i] == INF) continue;
        cnt[dist[i]]++;
    }
    if(flag == false)
    {
        sdist = dist;
        scnt = cnt;
    }
    else
    {
        tdist = dist;
        tcnt = cnt;
    }
}
signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int s, t; cin >> N >> M >> s >> t;
    s--, t--;
    for(int i = 0; i < M; i++)
    {
        int x, y; cin >> x >> y;
        if(x == y) continue;
        x--, y--;
        T[x].insert(y);
        T[y].insert(x);
    }
    search(s, t, 0);
    search(t, s, 1);
    L = sdist[t];
    ll ans = 0;
    for(int i = 0; i < N; i++)
    {
        int dist = L - 2 - sdist[i];
        if(dist < 0) continue;
        ans += tcnt[dist];
    }
    cout << ans << endl;
    return 0;
}